Answer
$f_x=\dfrac{y^2-x^2}{(x^2+y^2)^2}$
and
$f_y=\dfrac{-2xy}{(x^2+y^2)^2}$
Work Step by Step
Our aim is to take the first partial derivatives of the given function $f(x,y)$ with respect to $x$, by keeping $y$ as a constant.
$f_x=\dfrac{(x^2+y^2)-x \times (2x)}{(x^2+y^2)^2}=\dfrac{y^2-x^2}{(x^2+y^2)^2}$
Take the first partial derivatives of the given function $f(x,y)$ with respect to $y$, by keeping $x$ as a constant.
$f_y=\dfrac{(x^2+y^2)(0)-x \times (2y)}{(x^2+y^2)^2}=\dfrac{-2xy}{(x^2+y^2)^2}$
